Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 3 - Review Exercises - Page 136: 60



Work Step by Step

Note that $\dfrac{\pi}{6}$ is a special angle and $\sin{(\dfrac{\pi}{6})}=\frac{1}{2}$. RECALL: An angle and its reference angle have either the same trigonometric values or they differ only in signs. Since $s$ must be in $\left[\dfrac{3\pi}{2}, 2\pi\right]$, then the angle must terminate in Quadrant IV. Note that the angle of $\dfrac{11\pi}{6}$ is in Quadrant IV and its reference angle is $\dfrac{\pi}{6}$. Sine is negative in Quadrant IV. Thus, $\sin{(\frac{11\pi}{6})} = -\frac{1}{2}$ Therefore, if $\sin{s} = -\frac{1}{2}$ and $s$ is in $\left[\dfrac{3\pi}{2}, 2\pi\right]$, then $\color{blue}{s=\dfrac{11\pi}{6}}$.
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